Is anything known about its irreducible representations? In particular, how many nonequivalent irreps do exist? What are their dimensions? Can we construct explicit representation matrices for a given irrep?

Algebra over what ring? How exactly is "representation" defined if the algebra is non-associative? First observation: The algebra is commutative and $B$ is its neutral element.
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Johannes HahnJul 22 '11 at 20:15

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Second observation: If you make this associative, then $0 = A^2 C^2 = A(2B)C=2AC=4B$. Since $B$ is the multiplicative identity, this means that $4x=0$ for any x in your ring. If you don't make it associative, then I don't know what you mean by a representation.
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David SpeyerJul 22 '11 at 20:52

More over, in an associative version $0=AAC=A2B=2A$ holds. Since $\lbrace A,B,C\rbrace$ is assumed to be a basis, the characteristic must be two. Therefore the algebra is just $R[A,C]/(A^2,AC,C^2)$ with whatever base ring $R$ is choosen to be. The simple modules of this are just the simple $R$-modules with $A$ and $C$ acting as zero.
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Johannes HahnJul 22 '11 at 22:03

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To summarize the above comments: there is no (sensible) notion of representation for non-associative algebras, and therefore your questions doesn't make sense.
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André HenriquesJul 22 '11 at 22:42

I believe there is a notion of module over an algebra over an operad, or something along those lines, and I believe that specializing that definition to this case gives the following: if $A$ is a non-associative algebra (a magma internal to the category of $k$-vector spaces for some field $k$), then an $A$-module is a $k$-vector space $M$ equipped with a bilinear map $A \times M \to M$ satisfying... no axioms! Not a very interesting thing.
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Qiaochu YuanJul 23 '11 at 4:12

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Armed with Jacobson's Structure and representations of Jordan algebras, for example, you will be able to find the sensible notion of representation of your algebra (there are, in fact, a couple of sensible notions...) Using that, and since the algebra is of dimension $3$, which is hopefully small, one can possibly describe the irreducible ones.

Of course, this is more or less arbitrary: you have to choose in what sense you want to represent your algebra—this is what the comments above are hinting at—but, well, looking at it as a Jordan algebra makes this choice for you.

A representation of a Jordan algebra $(\mathbb J,\cdot)$ is a map $\rho:\mathbb J\to \mathrm{End}(V)$ such that $\rho(A\cdot B)=\rho(A)\rho(B)+\rho(B)\rho(A)$ for any $A,B\in \mathbb J$. Another option is to require the map $\rho$ to land in the subset of self-adjoint operators.
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André HenriquesJul 23 '11 at 20:12